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Find the normal approximation for the binomial probability \(P(x \geq 9),\) where \(n=13\) and \(p=0.7\) Compare this to the value of \(P(x \geq 9)\) obtained from Table 2.

Short Answer

Expert verified
The normal approximation for the binomial probability \(P(x \geq 9)\) is 0.6729. Comparison to the value from the Binomial distribution table may vary depending on the provided table.

Step by step solution

01

Calculate Mean and Standard Deviation

The mean \(\mu\) and standard deviation \(\sigma\) of a binomial distribution are calculated as follows: \(\mu = np\) and \(\sigma =\sqrt{np(1-p)}\). So, substituting the given values \(n=13\) and \(p=0.7\), we get \(\mu = 13*0.7 = 9.1\) and \(\sigma = \sqrt{13*0.7*0.3} = 1.345.\
02

Calculate the Z-score

The Z-score is computed as \((x - \mu) / \sigma\). Here, the number '9' for which the cumulative probability needs to found corresponds to the random variable 'x'. In this case, we need to apply the correction for continuity, hence \(x = 9 - 0.5 = 8.5.\) Thus, the Z-score is \((8.5 - 9.1) / 1.345 = -0.446.\
03

Use Z-table to find Normal Approximation for \(P(x \geq 9)\)

Finding the value of a Z-score of -0.446 on the Z-table, we get 0.3271. But this corresponds to \(P(x \leq 8.5)\). Since we want \(P(x \geq 9)\), we have to subtract this from 1. So, \(P(x \geq 9) = 1 - P(x \leq 8.5) = 1 - 0.3271 = 0.6729\)
04

Comparison with Binomial Table

The value of \(P(x \geq 9)\) from the Binomial distribution table has to be given (as the exercise refers to 'Table 2'). This should be directly compared with the normal approximation obtained in the previous step

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Distribution
The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials, each with the same probability of success.

For example, if you flip a coin (where the probability of getting heads is 0.5) 10 times, the distribution of the number of heads you obtain follows a binomial distribution. In general, if you have an experiment that can result in one of two outcomes (success or failure) and you repeat this experiment 'n' times independently, the probabilities of different possible numbers of successes follow the binomial distribution with parameters 'n' (number of trials) and 'p' (probability of success in a single trial).

This concept is vital when dealing with problems like the textbook question, where we calculate the likelihood of a certain number of successes out of a set number of trials.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. In the context of binomial distribution, it reflects the variability of the number of successes you can expect over a large number of trials.

It's calculated using the formula \[\sigma = \sqrt{np(1-p)}\], where 'n' is the number of trials, and 'p' is the probability of success. The standard deviation helps us understand the variability around the expected number of successes (the mean). A smaller standard deviation means the outcomes are tightly clustered around the mean, while a larger standard deviation indicates that the outcomes are spread out more widely.

In our specific exercise, we calculated the standard deviation to understand the distribution of the outcomes around the expected number of successes, which is crucial for accurately applying the normal approximation.
Z-score
A Z-score is a statistical measure that describes a value's relationship to the mean of a group of values. It's expressed in terms of standard deviations from the mean. If a Z-score is 0, it is at the mean; a positive Z-score indicates the number of standard deviations above the mean, and a negative Z-score indicates the number of standard deviations below the mean.

The Z-score is calculated using the formula \[ Z = \frac{(X - \mu)}{\sigma} \] where 'X' is the value we are comparing to the mean, '\mu' is the mean, and '\sigma' is the standard deviation. In the context of the normal approximation to the binomial distribution, the Z-score allows us to transition from a discrete to a continuous framework, making it possible to use the normal distribution to approximate binomial probabilities.

This calculation was integral to determining the likelihood of observing at least 9 successes in our exercise.
Continuity Correction
Continuity correction is an adjustment made when a discrete distribution (like the binomial) is approximated by a continuous distribution (like the normal). This adjustment is necessary because the binomial distribution is discrete but the normal distribution is continuous, which can lead to inaccuracies during approximation.

The typical adjustment involves adding or subtracting 0.5 from the discrete value before finding the corresponding Z-score. The reason for this is to account for the fact that, in a discrete distribution, each outcome is actually a range of values (since you can't have, say, 9.3 successes), but in a continuous distribution, every individual value is a separate instance. So, this correction improves the normal approximation of discrete data.

In our textbook example, we used continuity correction by adjusting the value of 'x' from 9 to 8.5 before calculating the Z-score, leading to a more accurate approximation of the probability.

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